\achapter

\answer $ T = 0 . 7 3 $秒

\answer 在竖直方向

\answer 相同

\answer $ \varphi _ { 0 } = 0 $

\answer (1) $ \dot{y} = A \omega \cos \omega t + 2 \omega \cos 2 \omega t$

\aindent $ \dot{y} = - A \omega ^ { 2 } \sin \omega t - 4 \omega ^ { 2 } \sin 2\omega t $

(2)不是

\addtocounter{answer}{2}
\answer (1) $ \nu = 1 . 5 $赫兹

(2) $ A = 2 . 2 $厘米

(3) $ \varphi _ { 0 } = 2 4 ^ \circ 3 7 ^ { \prime } = 0 . 4 3 $ 弧度

(4) $ x = \sqrt { 5 } \cos \left( 7 \sqrt { 2 } t + 0 . 4 3 \right) $

\answer $ T = 2 . 2 1 $秒

\answer $t=\pm \dfrac{\pi}{4 \omega}-\dfrac{\varphi_{0}}{\omega}+\dfrac{n \pi}{\omega}, n=0, \pm 1, \pm 2, \ldots$

\answer $ T = 0 . 7 7 $秒

\answer $ m = 1 . 6 $公斤

\answer (1) 振幅$ A \geqslant 2 4 . 8 3 $厘米时

(2) $ \nu _ {\text{max}} = 2 . 2 3 $赫兹

\answer $ A _ { \text{max} } = 3 . 1 $厘米

\answer $ 120^{\circ} $

\answer (1) $ 14 $牛顿/米

(2) $ M = 7 9 $克

\answer $ k _ { 1 } = \dfrac { n + 1 } { n } k , k _ { n } = \left( n + 1 \right) k $

\answer $ \nu = \dfrac { 1 } { 2 \uppi } \sqrt { \dfrac { k _ { 1 } + k _ { 2 } } { m } } $

\answer $ C $点为质心，不动，图\ref{fig:07.17} 中有$ m _ { 2 } l _ { 2 } = m _ { 1 } l _ { 1 } $

% 391.jpg
$ m _ 2 $对质心作简谐振动，振动方程为

$x_{2}=l_{2}+(l_{2}^{\prime}-l_{2}) \cos \omega_{2} t,\quad \omega_{2}=\sqrt{\dfrac{k(m_{1}+m_{2})}{m_{1} m_{2}}}$

$ m_1 $对质心作简谐振动，振动方程为

$x_{1}=-l_{1}+(l_{1}^{\prime}-l_{1}) \cos \omega_{1} t, \quad \omega_{2}=\omega_{2}$

\answer $ C $点作自由落体运动，$ m _ { 1 }, m _ 2 $相对$ C $的运动同上题

\answer (1) $ T = 2 \uppi \sqrt { \dfrac { l _ { 1 } ^ 2 + l _ { 2 } ^ 2 } { ( l _ { 1 } - l _ { 2 } ) g } } = 1 . 7$秒

(2) $ l _ { 0 } = 7 5 $厘米

\stepcounter{answer}
\answer $ 75 $公里/时

\answer $ Q \approx 7,0 0 0 $

\answer $ Q \approx 1 3 $
